Question

A ball is thrown vertically upward with speed $70m/s$. After $1$s, another ball is dropped from a height $185$m in the same vertical line. At what height the balls collide?

• A. $150$m
• B. $160$m
• C. $168$m
• D. $165$m

Answer 1

Answer: (D)

$165$m

Here, the second ball is dropped $1$s after the projection of the first. Let the balls meet at height h after t seconds, after the projection of first ball.

Second ball will take time $\left(t\right)t-1$, because it was dropped after $1$s.

Ball I, h$=70t-\frac{1}{2}g{t}^2$ ...........(i)

Ball II, $185-h=\frac{1}{2}g(t-1{)}^2$ ............(ii)

On adding Eqs. (i) and (ii), we get

$185=70t-\frac{1}{2}g[t^2-(t-1{)}^2]$

$=70t-\frac{1}{2}g(2t-1)\left(1\right)$

$=70t-10t+5$

$60t=180$

$\Rightarrow t=3s$

$h=70t-\frac{1}{2}g{t}^2$ [from Eq. (i)]

$=70\times 3-\frac{1}{2}\times 10\times 3^2$

$=210-45=165$m.